a_507_bc wrote: Let $ABC$ be an acute triangle with $AB>AC$. The internal angle bisector of $\angle BAC$ meets $BC$ at $D$. Let $O$ be the circumcenter of $ABC$ and let $AO$ meet $BC$ at $E$. Let $J$ be the incenter of triangle $AED$. Show that if $\angle ADO=45^{\circ}$, then $OJ=JD$. By angle chasing $\angle DOA=135^\circ+\frac{1}{2}\beta-\frac{1}{2}\gamma=\angle DEA$. So $A,D,J,O$ lie on a circle and since $J$ lies on the angle bisector of $\angle OAD$ it is the $A$-south pole in the triangle $ADO$. Thus $OJ=JD$.

This problem was proposed by Burii.

Seems that we don't even need cyclic quadrilaterals, only the fact that angles in a triangle sum to $180^\circ$: We have $\angle ADB = \gamma + \frac{\alpha}{2}$ and therefore $\angle ODE = \gamma + \frac{\alpha}{2} - 45^\circ$. Also, $\angle DEO = \angle EBA + \angle BAO =\beta + 90 ^\circ - \gamma$. This implies that $\angle DEO + 2 \cdot \angle ODE = 180^\circ$ and thus $\angle ODE = \angle EOD$. Hence, $ED=EO$, which means that $D, O$ are symmetric wrt $EJ$ and therefore $DJ=JO$. So in fact we only used that $J$ is on the angle bisector of $\angle DEA$, not necessarily the incenter.